package leetcode.editor.cn;

/**
 * @id: 79
 * @title: 单词搜索
 */
 
//给定一个 m x n 二维字符网格 board 和一个字符串单词 word 。如果 word 存在于网格中，返回 true ；否则，返回 false 。 
//
// 单词必须按照字母顺序，通过相邻的单元格内的字母构成，其中“相邻”单元格是那些水平相邻或垂直相邻的单元格。同一个单元格内的字母不允许被重复使用。 
//
// 
//
// 示例 1： 
//
// 
//输入：board = [['A','B','C','E'],['S','F','C','S'],['A','D','E','E']], word = 'AB
//CCED'
//输出：true
// 
//
// 示例 2： 
//
// 
//输入：board = [['A','B','C','E'],['S','F','C','S'],['A','D','E','E']], word = 'SE
//E'
//输出：true
// 
//
// 示例 3： 
//
// 
//输入：board = [['A','B','C','E'],['S','F','C','S'],['A','D','E','E']], word = 'AB
//CB'
//输出：false
// 
//
// 
//
// 提示： 
//
// 
// m == board.length 
// n = board[i].length 
// 1 <= m, n <= 6 
// 1 <= word.length <= 15 
// board 和 word 仅由大小写英文字母组成 
// 
//
// 
//
// 进阶：你可以使用搜索剪枝的技术来优化解决方案，使其在 board 更大的情况下可以更快解决问题？ 
// Related Topics 数组 回溯 矩阵 
// 👍 1148 👎 0

public class P79WordSearch {
    public static void main(String[] args) {
        Solution solution = new P79WordSearch().new Solution();
        // todo
        char[][] chars = new char[][] {
                new char[]{'A','B','C','E'},
                new char[]{'S','F','C','S'},
                new char[]{'A','D','E','E'}};
//        [["a","a","b","a","a","b"],["a","a","b","b","b","a"],["a","a","a","a","b","a"],["b","a","b","b","a","b"],["a","b","b","a","b","a"],["b","a","a","a","a","b"]]
//        bbbaabbbbbab
//        char[][] chars = new char[][] {
//                new char[]{'a','a','a','a'},
//                new char[]{'a','a','a','a'},
//                new char[]{'a','a','a','a'}
//        };
        final boolean exist = solution.exist(chars, "ABCB");
        System.out.println(exist);
    }
    
//leetcode submit region begin(Prohibit modification and deletion)
class Solution {
    // 上、右、下、左
    private int[] drow = new int[]{-1, 0, 1, 0};
    private int[] dcol = new int[]{0, 1, 0, -1};

    public boolean exist(char[][] board, String word) {
        final int n = board.length;
        final int m = board[0].length;
        if (word.length() > n * m) {
            return false;
        }
        boolean[][] memo = new boolean[n][m];
        for (int i = 0; i < n; ++i) {
            for (int j = 0; j < m; ++j) {
                final boolean result = findWord(board, memo, i, j, 0, word);
                if (result) {
                    return true;
                }
            }
        }
        return false;
    }

    // 不能直接构建对比串，这样每次都要字符串对比，多了一次 O(word.len)遍历
    private boolean findWord(char[][] board,
                          boolean[][] memo,
                          int r,
                          int c,
                          int k,
                          String target) {
        if (target.charAt(k) != board[r][c]) {
            return false;
        } else if (k == target.length() - 1) {
            return true;
        }
        memo[r][c] = true;
        for (int i = 0; i < 4; ++i) {
            int nextRow = r + drow[i];
            int nextCol = c + dcol[i];
            if (nextRow < 0 || nextRow >= board.length
                    || nextCol < 0 || nextCol >= board[0].length
                    || memo[nextRow][nextCol]) {
                continue;
            }
            boolean result = findWord(board, memo, nextRow, nextCol, k + 1, target);
            if (result) {
                return true;
            }
        }
        memo[r][c] = false;
        return false;
    }

}
//leetcode submit region end(Prohibit modification and deletion)


}